The molality of 49% by volume of `H_(2)SO_(4)` solution having density 1.49 g/mL is……

Explanation

To solve the problem of finding the molality of a 49% by volume `H2SO4` solution with a density of 1.49 g/mL, we will follow these steps:

### Step 1: Assume a Volume of Solution
Assume a volume of 1000 mL (1 L) for easier calculations.

### Step 2: Calculate the Mass of the Solution
Using the density of the solution, we can calculate the mass of the solution:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.49 \, \text{g/mL} \times 1000 \, \text{mL} = 1490 \, \text{g} \]

### Step 3: Calculate the Weight of the Solute
Since the solution is 49% by volume, we can find the weight of the solute (H2SO4):
\[ \text{Weight of solute} = 49\% \text{ of 1000 mL} = 490 \, \text{g} \]

### Step 4: Calculate the Weight of the Solvent
Now, we can calculate the weight of the solvent:
\[ \text{Weight of solvent} = \text{Total weight of solution} - \text{Weight of solute} = 1490 \, \text{g} - 490 \, \text{g} = 1000 \, \text{g} \]
Convert this to kilograms:
\[ \text{Weight of solvent} = 1000 \, \text{g} = 1 \, \text{kg} \]

### Step 5: Calculate the Number of Moles of Solute
Next, we need to calculate the number of moles of the solute (H2SO4). The molecular weight of H2SO4 is 98 g/mol:
\[ \text{Number of moles of solute} = \frac{\text{Weight of solute}}{\text{Molecular weight of solute}} = \frac{490 \, \text{g}}{98 \, \text{g/mol}} = 5 \, \text{moles} \]

### Step 6: Calculate the Molality
Finally, we can calculate the molality (m):
\[ \text{Molality} = \frac{\text{Number of moles of solute}}{\text{Weight of solvent in kg}} = \frac{5 \, \text{moles}}{1 \, \text{kg}} = 5 \, \text{mol/kg} \]

### Final Answer
The molality of the 49% by volume `H2SO4` solution is **5 mol/kg**.

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